Answer:
Q = - 1.95 kCal
Solution:
The equation used for this problem is as follow,
Q = m Cp ΔT
Where'
Q = Heat = ?
m = mass = 85 g
Cp = Specific Heat Capacity = 4.18 J.g⁻¹.°C⁻¹
ΔT = Change in Temperature = 67 °C - 90 °C = - 23 °C
Putting values in eq. 1,
Q = 85 g × 4.18 J.g⁻¹.°C⁻¹ × - 23 °C
Q = - 8172 J
Q = - 1.95 kCal